Question

Consider a random sample of n = 10 observations that
yield a mean of 64.46. Assume that the population is normal with standard
deviation = 4:5.
(a) Construct a 95% C.I. for .
(b) Construct a 98% C.I. for .
(c) Compare the precision of the two intervals.

Answer 1

Solution :

Given that,

a ) = 64.46

= 4.5

n = 10

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Margin of error = E = Z_/2* (/n)

= 1.960 * (4.5 / 10 ) = 2.79

At 95% confidence interval estimate of the population mean is,

- E < < + E

64.46 - 2.79 < < 64.46 + 2.79

61.67 < < 67.25

(61.67 , 67.25)

b ) = 64.46

= 4.5

n = 10

At 98% confidence level the z is ,

  = 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z_/2* (/n)

= 2.326 * (4.5 / 10 ) = 3.31

At 98% confidence interval estimate of the population mean is,

- E < < + E

64.46 - 3.31 < < 64.46 + 3.31

61.15 < < 67.77 )

(61.15 , 67.77 )

c ) The 98% confidence inerval wider than 95 % confidence inerval.